Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <algorithm>

int n;
double dp[3000];

int main(){
  scanf("%d", &n);
  dp[0] = 1;
  for(int i = 1; i <= n; i++){
    double p;
    scanf("%lf", &p);
    for(int j = i; j >= 0; j--){
      dp[j+1] += dp[j] * p;
      dp[j] *= 1-p;
    }
  }
  double ans = 0;
  for(int i = n/2+1; i <= n; i++) ans += dp[i];
  printf("%.16f\n", ans);

  return 0;
}